Hobby Electronics Basics Newbie needs help with basics

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Subject Author Date
Newbie needs help with basics hdjim69 03-16-06
Posted by Phil Allison on March 16, 2006, 7:32 pm
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"James Douglas"

>
> Jeesh, go to your doctor and get the pine cone out of your ass!


** FUCK OFF - IDIOT !!

The OP was an arrogant moron.

YOU are even WORSE !




......... Phil







Posted by Greg Neill on March 16, 2006, 11:38 pm
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>
> "James Douglas"
>
> >
> > Jeesh, go to your doctor and get the pine cone out of your ass!
>
>
> ** FUCK OFF - IDIOT !!
>
> The OP was an arrogant moron.
>
> YOU are even WORSE !

Bye Bye potty mouth:

*PLONK* *FLUSH*



Posted by hdjim69 on March 16, 2006, 7:51 am
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Hey Phil, chill out !! The link should have taken right to the
problem. It's Activity 1 exercise 2.


Posted by Phil Allison on March 16, 2006, 8:02 am
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> Hey Phil, chill out !!


** WELL FUUUCK YOU !!!

Mr Googlegroups, anonymous "hotmail" SHITHEAD !!!!

SCUM of THE EARTH TROLL !!!



> The link should have taken right to the
> problem.



** Would anyone complain if it FUCKING DID ??

YOU fucking ASSHOLE !!



....... Phil





Posted by Greg Neill on March 16, 2006, 7:50 am
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> Hi, I'm trying to learn electronics, I'm right at the beginning,
> just learned Ohms law and Watts law. I understand both. I understand
> how to calculate both. However, I went to this tutorial on
> Electronicsworkbench.com and I'm confused about how they come up with
> these numbers.
>
> In this example there is a series circuit with a 15v battery, a 10w/12v
> light bulb with a voltmeter reading the voltage across the bulb and you
> have to choose between 3 different resistors that will enable the bulb
> to light - 1ohm, 100ohm and 5ohm. If you connect the 1ohm, the
> voltmeter reads 15volts. How can that be ?

If the current is too high the bulb will burn out and the
full supply voltage will appear across the failed (opened)
bulb.

> In a series circuit each
> load has some resistance and the sum of all the loads must equal the
> source. So, what about the resistor ? It has to drop some voltage but
> according to this, it's dropping zero.

When the bulb is burnt out, no current will flow so there
will be no voltage drop across the resistor.

> And if you connect the 100
> ohm resistor, the voltmeter reads 1.88v. How did they come up with
> that number ?

If the lamp has a rating of 10W at 12V, you can calculate a
resistance for the bulb from P = V^2/R. You then have an
equivalent circuit consisting of two resistors in series. Note
that real light bulbs have different resistance for different
operating points (resistance depends upon the temperature of the
filament). In this simplified example, they're probably ignoring
this little detail.

> And the 5ohm reads 12v across the voltmeter. But
> according to Ohms law, E = IxR, if the resistor is 5ohms and the source
> is 15v, then 15/5 = 3amps, that part I get but they are saying it's
> dropping 3 volts ?? huh? Again, how did they get that value ?

You forgot to include the resistance of the bulb, or alternativly,
if you assume that the bulb is operating at its correct operating
point and dropping 12V at 0.833A (for a power dissipation of 10W),
that leaves (15 - 12)V = 3V to drop across the resistor.



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